Chemical Bonding and Molecular Structure Test -9

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Chemical Bonding and Molecular Structure Test -9

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Weekly Quiz Competition

Question 1
1 / -0

Isostructural species are those which have the same shape and hybridisation. Among the given species identify the isostructural pairs.

A
$ [NF_3\, and\, BF_3]$

B
$[BF^-_4\, and\, NH^+_4]$

C
$[BCl_3\,and\, BrCl_3]$

D
$[NH_3\,and\, NO^-_3]$

Solution

Both the molecules have $sp ^3$ hybridisation and shape is tetrahedral.
Question 2
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Polarity in a molecule and hence the dipole moment depends primarily on electronegativity of the constituent atoms and shape of a molecule. Which of the following has the highest dipole moment?

A
$CO_2$

B
$HI$

C
$H_2O$

D
$SO_2$

Solution

Oxygen is highly electronegative and because of two lone pair on oxygen, the dipole moment is high.
Question 3
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The types of hybrid orbitals of nitrogen in $NO_2^+$, $NO^-_3$ and $NH^+_4$ respectively are expected to be

A
$sp, sp^3\, and\, sp^2$

B
$sp, sp^2\, and\, sp^3$

C
$sp^2, sp\, and\, sp^3$

D
$sp^3, sp^2\, and\, sp$

Solution

The hybridization in $NO_2^+$ is sp and in $NO^-_3$ $sp^2$ and in $NH^+_4$, it is $sp ^3$
Question 4
1 / -0

Hydrogen bonds are formed in many compounds e.g., $H_2O$, $HF$, $NH_3 $. The boiling point of such compounds depends to a large extent on the strength of hydrogen bond and the number of hydrogen bonds. The correct decreasing order of the boiling points of above compounds is :

A
$HF > H_2O > NH_3$

B
$H_2O > HF > NH_3$

C
$NH_3 > HF > H_2O$

D
$NH_3 > H_2O > HF$

Solution

Strength of hydrogen bond depends on electronegativity, size of atom and number of hydrogen bonds per mol. F is most electronegative but, in $H_2O$, the no. of H-Bonds are more and size is smaller than N.
Question 5
1 / -0

In $PO^{3-}_4$ ion the formal charge on the oxygen atom of P-O bond is

A
+1

B
-1

C
-0.75

D
+0.75

Solution

The formal charge of an atom in a polyatomic molecule or ion may be defined as the difference between the number of valence electrons of that atom in an isolated or free state and the number of electrons assigned to that atom in the Lewis structure.

Formal charge of the atom in the molecule or ion = (Number of valence electron in the free atom) - (Number of lone pair electrons + $1\over2$ number of bonding electrons)

So formal charge on each O-atom

= 6 - (6+ $1\over2$ x 2) = 6 - 7 = -1
Question 6
1 / -0

Which of the following have identical bond order?

(i) $CN^-$

(ii) $NO^+$

(iii) $O^-_2$

(iv) $O^{2-}_2$

A
(i),(ii)

B
(ii),(iii)

C
(iii),(iv)

D
(i),(iv)

Solution

In both the species number of electrons in $N_b$ = number of electrons in $N_{ab}$

Question 7

1 / -0

Match the species in Column I with the type of hybrid orbitals in Column II

Column I	Column II
(i) $SF_4$	(a) $sp^3d^2$
(ii) $IF_5$	(b) $d^2sp^3$
(iii) $NO^+_2$	(c) $sp^3d$
(iv) $NH^+_4$	(d) $sp^3$
	(e) sp

(i) → (b) (ii) → (a) (iii) → (e) (iv) → (c)

(i) → (c) (ii) → (a) (iii) → (e) (iv) → (b)

(i) → (c) (ii) → (a) (iii) → (e) (iv) → (d)

(i) → (d) (ii) → (c) (iii) → (a) (iv) → (e)

Solution

Column I	Column II
(i) $SF_4$	S is surrounded by 4 bond pairs and 1 lone pair ($sp^ 3d$ hybridization).
(ii) $IF_5$	I is surrounded by 5 bond pairs and 1 lone pair ($sp^ 3d ^2$ hybridization).
(iii) $NO^+_2$	N has 2 bond pairs and no lone pair (sp hybridization).
(iv) $NH^+_4$	N has 4 bond pairs and no lone pair ($sp ^3$ hybridization).

Question 8
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In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion (A): Sodium chloride formed by the action of chlorine gas on sodium metal is a stable compound.

Reason (R): This is because sodium and chloride ions acquire octet in sodium chloride formation.

A
A and R both are correct, and R is the correct explanation of A

B
A and R both are correct, but R is not the correct explanation of A

C
A is true but R is false

D
A and R both are false

Solution

$\underset{2,8,1}{Na} + \underset{2,8,7}{Cl}\rightarrow \underset{(2,8)(2,8,8)}{NaCl}$

In NaCl, Sodium and Chloride ions have complete octet hence, NaCl is stable compound.
Question 9
1 / -0

Which of the following attain the linear structure:

(i) $BeCl_2$

(ii) $NCO^ +$

(iii) $NO_2$

(iv) $CS_2$

A
(i),(ii)

B
(ii),(iii)

C
(iii),(iv)

D
(i),(iv)

Solution

In both, the central atom is sp hybridized.
Question 10
1 / -0

Which of the following species has tetrahedral geometry?

A
$BH_4^-$

B
$NH_2^-$

C
$CO_3^{2-}$

D
$H_3O^+$

Solution

Boron is surrounded by 4 bonded pairs only

In $BH_4^-$ = no. of bond pair = 4

no. of lone pair = 0

$sp^3$ hybridised, so have tetrahedral geometry

$NH_2^-$ = V - shape $H_3O^+$ = Pyramidal

$CO_3^{2-}$ = triangular planar
Question 11
1 / -0

CO is isoelectronic with

(i) $NO ^+$

(ii) $N_2$

(iii) $SnCl_2$

(iv) $NO^-_2$

A
(ii),(iv)

B
(i),(ii)

C
(i),(iii)

D
(ii),(iii)

Solution

In both the species the number of electrons are same but the number of protons are different.

Question 12

1 / -0

Match the species in Column I with the geometry/shape in Column II.

Column I	Column II
(i) $H_3O^+$	(a) Linear
(ii) HC $\equiv$ CH	(b) Angular
(iii) $ClO^-_2$	(c) Tetrahedral
(iv) $NH^+_4$	(d) Trigonal bipyramidal
	(e) Pyramidal

(i) → (e) (ii) → (a) (iii) → (b) (iv) → (c)

(i) → (e) (ii) → (a) (iii) → (c) (iv) → (d)

(i) → (c) (ii) → (a) (iii) → (b) (iv) → (e)

(i) → (e) (ii) → (b) (iii) → (d) (iv) → (c)

Solution

Column I	Column II
(i) $H_3O^+$	Oxygen has 3 bond pairs and 1 lone pair (pyramidal shape)
(ii) HC $\equiv$ CH	linear as sp hybridized
(iii) $ClO^-_2$	Cl has 2 bond pairs and 2 lone pairs (Angular shape)
(iv) $NH^+_4$	N has 4 bond pairs and no lone pair (Tetrahedral)

Question 13
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Which of the following species have the same shape?

(i) $CO_2$

(ii) $CCl_4$

(iii) $O_3$

(iv) $NO^-_2$

A
(i),(ii)

B
(ii),(iii)

C
(iii),(iv)

D
(i),(iv)

Solution

Central atom in both the species has same hybridized state and same geometry.
Question 14
1 / -0

Which molecule/ion out of the following does not contain unpaired electrons?

A
$N^+_2$

B
$O_2$

C
$O^{2-}_2$

D
$B_2$

Solution

In $O^{2-}_2$ antibonding $P_{nx}$ = $P_{ny}$ orbitals have paired electrons.
Question 15
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In which of the following substances will hydrogen bond be strongest?

A
$HCl$

B
$H_2O$

C
$HI$

D
$H_2S$

Solution

Oxygen is more electronegative as compared to others and its size is small.
Question 16
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Match the species in Column I with the bond order in Column II.

Column I Column II

(i) N0 (a) 1.5

(ii) CO (b) 2.0

(iii) $O^-_2$ (c) 2.5

(iv) $O_2$ (d) 3.0

A
(i) → (b) (ii) → (c) (iii) → (d) (iv) → (a)

B
(i) → (b) (ii) → (d) (iii) → (a) (iv) → (c)

C
(i) → (c) (ii) → (d) (iii) → (b) (iv) → (a)

D
(i) → (c) (ii) → (d) (iii) → (a) (iv) → (b)
Question 17
1 / -0

What is Bond formation?

A
always exothermic

B
always endothermic

C
neither exothermic nor endothermic

D
Sometimes exothermic and sometimes endothermic

Solution

Chemical bond formation is always an exothermic process and chemical bond breaking is an endothermic process.
Hence, option A is correct.
Question 18
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An _____ bond is a bond in which electrons are transferred.

A
Ionic

B
Covalent

C
Double

D
Triple

Solution

Ionic bond is the bond in which electrons are transferred.
Question 19
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Which one of these is weakest ?

A
ionic bond

B
covalent bond

C
metallic bond

D
van der Waal forces

Solution

Out of the given options, van der waal forces are the weakest.
Hence, option D is correct.
Question 20
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The number of sigma and $$\pi$$ bonds in 1-butene-3yne are-

A
$$5$$ sigma and $$5$$ pi

B
$$7$$ sigma and $$3$$ pi

C
$$8$$ sigma and $$2$$ pi

D
$$6$$ sigma and $$4$$ pi

Solution
Question 21
1 / -0

Compound having maximum number of bonded pairs of electrons in its molecule is:

A
Ethane

B
Ammonia

C
Sulphur hexafluoride

D
Bromine Pentafluoride

Solution

(a) Ethane has $$6 \:C-H$$ bonds and $$1 \:C-C$$ bond.
(b) Ammonia has $$3\: N-H$$ bonds.
(c) Sulphur hexafluoride has $$6\: S-F$$ bonds.
(d) Bromine pentafluoride has $$5\: Br-F$$ bonds.
Question 22
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Molecule which contains only bonded pairs of electrons on the central atom is:

A
$$H_{2}O$$

B
$$NH_{3}$$

C
$$BeCl_{2}$$

D
$$BF_{3}$$

Solution

$$Be$$ has $$2$$ electrons in the valence shell (it belongs to group $$2$$ of the periodic table).

Now both the electrons are involved in bond formation with both the $$Cl$$ atoms. It comprises only bonded pair of electrons.

So, there are no lone pairs in the $$Be$$ atom.

Hence, option C is correct.
Question 23
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Number of bonded electrons in ethane molecule are:

A
$$7$$

B
$$12$$

C
$$10$$

D
$$14$$

Solution

Ethane consists of $$6\:C-H$$ bonds and $$ 1\: C-C$$ bond. Total number of bonds is $$7$$. Each bond is made up of two electrons. Thus, total bonded electrons are $$7\times2$$ = $$14$$.
Question 24
1 / -0

The number of electron bond pairs involved in the formation of hydrogen cyanide molecule are:

A
two

B
eight

C
three

D
four

Solution

$$HCN$$ contains $$4$$ bonds ( $$1\: C-H$$ single bond and $$1$$ triple bond between $$C$$ and $$N$$). Thus, total bond pairs are $$4$$.
Question 25
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In which of the following molecule, the central atom has three lone pairs of electrons?

A
Ammonia

B
Xenon difluoride

C
Chlorine trifluoride

D
Hydrogen sulphide

Solution

Xenon difluoride is $$XeF_{2}$$. $$Xe$$ is a noble gas. So, it has $$8 $$ electrons in the valence shell. Out of which, $$2$$ are involved in formation of covalent bond with the $$2 \:F$$ atoms. So, $$6$$ electrons are left as unbonded. Therefore, lone pairs = half of $$6$$ = $$3$$.
Question 26
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Molecule which contains $$4$$ bonded pairs and $$2$$ lone pairs of electrons on the central atom is:

A
$$XeF_{2}$$

B
$$CO_{2}$$

C
$$XeF_{4}$$

D
$$SF_{6}$$

Solution

$$Xe$$ is a noble gas. It has $$8$$ electrons in its valence shell. Now, it is bonded to $$4\: F$$ atoms.

So, $$4$$ of its electrons are involved in bonding.

Hence, bond pairs = $$4$$. Remaining $$4$$ are unbonded electrons. So, lone pairs = $$4 / 2 = 2$$.

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Column I	Column II
(i) \(SF_4\)	(a) \(sp^3d^2\)
(ii) \(IF_5\)	(b) \(d^2sp^3\)
(iii) \(NO^+_2\)	(c) \(sp^3d\)
(iv) \(NH^+_4\)	(d) \(sp^3\)
	(e) sp

Column I	Column II
(i) \(SF_4\)	S is surrounded by 4 bond pairs and 1 lone pair (\(sp^ 3d\) hybridization).
(ii) \(IF_5\)	I is surrounded by 5 bond pairs and 1 lone pair (\(sp^ 3d ^2\) hybridization).
(iii) \(NO^+_2\)	N has 2 bond pairs and no lone pair (sp hybridization).
(iv) \(NH^+_4\)	N has 4 bond pairs and no lone pair (\(sp ^3\) hybridization).

Column I	Column II
(i) \(H_3O^+\)	(a) Linear
(ii) HC \(\equiv\) CH	(b) Angular
(iii) \(ClO^-_2\)	(c) Tetrahedral
(iv) \(NH^+_4\)	(d) Trigonal bipyramidal
	(e) Pyramidal

Column I	Column II
(i) N0	(a) 1.5
(ii) CO	(b) 2.0
(iii) \(O^-_2\)	(c) 2.5
(iv) \(O_2\)	(d) 3.0

Chemical Bonding and Molecular Structure Test -9

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Chemical Bonding and Molecular Structure Test -9

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Question 1

\( [NF_3\, and\, BF_3]\)

\([BF^-_4\, and\, NH^+_4]\)

\([BCl_3\,and\, BrCl_3]\)

\([NH_3\,and\, NO^-_3]\)

Solution

Question 2

\(CO_2\)

\(HI\)

\(H_2O\)

\(SO_2\)

Solution

Question 3

\(sp, sp^3\, and\, sp^2\)

\(sp, sp^2\, and\, sp^3\)

\(sp^2, sp\, and\, sp^3\)

\(sp^3, sp^2\, and\, sp\)

Solution

Question 4

\(HF > H_2O > NH_3\)

\(H_2O > HF > NH_3\)

\(NH_3 > HF > H_2O\)

\(NH_3 > H_2O > HF\)

Solution

Question 5

+1

-1

-0.75

+0.75

Solution

Question 6

(i),(ii)

(ii),(iii)

(iii),(iv)

(i),(iv)

Solution

Question 7

(i) → (b) (ii) → (a) (iii) → (e) (iv) → (c)

(i) → (c) (ii) → (a) (iii) → (e) (iv) → (b)

(i) → (c) (ii) → (a) (iii) → (e) (iv) → (d)

(i) → (d) (ii) → (c) (iii) → (a) (iv) → (e)

Solution

Question 8

A and R both are correct, and R is the correct explanation of A

A and R both are correct, but R is not the correct explanation of A

A is true but R is false

A and R both are false

Solution

Question 9

(i),(ii)

(ii),(iii)

(iii),(iv)

(i),(iv)

Solution

Question 10

\(BH_4^-\)

\(NH_2^-\)

\(CO_3^{2-}\)

\(H_3O^+\)

Solution

Question 11

(ii),(iv)

(i),(ii)

(i),(iii)

(ii),(iii)

Solution

Question 12

(i) → (e) (ii) → (a) (iii) → (b) (iv) → (c)

(i) → (e) (ii) → (a) (iii) → (c) (iv) → (d)

(i) → (c) (ii) → (a) (iii) → (b) (iv) → (e)

(i) → (e) (ii) → (b) (iii) → (d) (iv) → (c)

Solution